What is the integral of sqrt(1+1/x) ?

The integral of sqrt(1+1/x) is sqrt(x)sqrt(x+1)+1/2 ln|sqrt((x+1)/x)+1|-1/2 ln|sqrt((x+1)/x)-1|+C

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integral of sqrt(1+1/x)

\int 1 + \frac{1}{x} d x

x x + 1 + \frac{1}{2} ln \frac{x + 1}{x} + 1 - \frac{1}{2} ln \frac{x + 1}{x} - 1 + C

Solution steps

\int 1 + \frac{1}{x} d x

Apply u-substitution

: \int - \frac{1 + u}{u ^{2}} d u

= \int - \frac{1 + u}{u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1 + u}{u ^{2}} d u

Apply Integration By Parts

: - \frac{1 + u}{u} - \int - \frac{1}{2 u 1 + u} d u

= - (- \frac{1 + u}{u} - \int - \frac{1}{2 u 1 + u} d u)

\int - \frac{1}{2 u 1 + u} d u = \frac{1}{2} ln 1 + u + 1 - \frac{1}{2} ln 1 + u - 1

= - (- \frac{1 + u}{u} - (\frac{1}{2} ln 1 + u + 1 - \frac{1}{2} ln 1 + u - 1))

Substitute back

u = \frac{1}{x}

= - - \frac{1 + \frac{1}{x}}{\frac{1}{x}} - (\frac{1}{2} ln 1 + \frac{1}{x} + 1 - \frac{1}{2} ln 1 + \frac{1}{x} - 1)

Simplify

- - \frac{1 + \frac{1}{x}}{\frac{1}{x}} - (\frac{1}{2} ln 1 + \frac{1}{x} + 1 - \frac{1}{2} ln 1 + \frac{1}{x} - 1) : x x + 1 + \frac{1}{2} ln \frac{x + 1}{x} + 1 - \frac{1}{2} ln \frac{x + 1}{x} - 1

= x x + 1 + \frac{1}{2} ln \frac{x + 1}{x} + 1 - \frac{1}{2} ln \frac{x + 1}{x} - 1

Add a constant to the solution

= x x + 1 + \frac{1}{2} ln \frac{x + 1}{x} + 1 - \frac{1}{2} ln \frac{x + 1}{x} - 1 + C

\frac{d}{d x} (6 e^{5 x} + e^{- x^{2}})

x^{2}, 4 x - x^{2}

\int \frac{1}{25 e ^{- 4 x} + e ^{4 x}} d x

f (x) = (5 x^{6} + 2 x^{3})^{4}

\int \frac{1}{- x - 1} d x

What is the integral of sqrt(1+1/x) ?
The integral of sqrt(1+1/x) is sqrt(x)sqrt(x+1)+1/2 ln|sqrt((x+1)/x)+1|-1/2 ln|sqrt((x+1)/x)-1|+C